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We want to design a synchronous counter that counts the sequence 0-1-0-2-0-3 and then repeats. The minimum number of J-K flip-flops required to implement this counter is
- 1
- 2
- 3
- 4
Correct answer: 3
Solution
The sequence 0-1-0-2-0-3 has six steps and the three occurrences of 0 are different states (they lead to different next outputs 1,2,3), so the counter needs 6 distinct states. Six states require ceil(log2 6)=3 flip-flops, so 3 J-K flip-flops are needed (index 2).
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